Question

2 mol of an ideal gas expanded reversibly and isothermally from 10 L to 100 L at 27°C. Heat involved in this process will be​

Answer 1

Given:

2 mol of an ideal gas expanded reversibly and isothermally from 10 L to 100 L at 27°C.

To find:

Heat involved in this process?

Calculation:

First of all, in isothermal process , ∆U (or change in internal energy) is zero , so we can say:

1st Law of Thermodynamics:

$\sf\Delta Q =\Delta U + W$

$\sf \implies\Delta Q =0 + W$

$\sf \implies\Delta Q =W$

$\sf \implies\Delta Q =nRT \ln \bigg(\dfrac{V_{f}}{V_{i}} \bigg)$

$\sf \implies\Delta Q =2.303nRT \log \bigg(\dfrac{V_{f}}{V_{i}} \bigg)$

$\sf \implies\Delta Q =2.303 \times 2 \times 8.314 \times 300 \times \log \bigg(\dfrac{100}{10} \bigg)$

$\sf \implies\Delta Q =2.303 \times 2 \times 8.314 \times 300 \times 1$

$\sf \implies\Delta Q =11488.2 \: joule$

$\sf \implies\Delta Q =11.4 \: kilojoule$

So, heat change is 11.4 kJ.